A student says that he had applied a force F=-k√x on a particle and the particle moved in simple harmonic motion.He refuses to tell whether k is contant or not.Assume that he has worked only with positive x and no other force acted on the particle.
(A)As x increases k increases
(B)Ax x increases k decreases
(C)As x increases k remains constant
(D)The motion cannot be simlpe harmonic
My book has given the correct choice as (A).A motion is simple harmonic only when the displacement is directly proportional to the displacement.But in the given question foce is directly proportional to the square root of the displacement.Also he has worked only with positive x.So the motion cannot be a simple harmonic motion.So option (D) should be correct.Am I correct?
1yeah u are but luk from the frame of the student for him if k increases with x then the motion is shm he hastn tried negative x so for him it is shm
1K is a force or spring constant.It is always constant.If it changes then it means that the time period of the motion also changes which also means that angular frequency of the motion also changes.Therefore, I think the motion is not simple harmonic.
1could it be that the book wants
k= K √x since x>0 and K= spring constant
so that F=-kx so that the particle performs S.H.M
1If k=K√x then it means that F=-k,which is not the equation of the simple harmonic motion.Also if k√x=K
where K=spring constant then how can k be a constant
1see don't always take k to be a SPRING CONSTANT..this is your first mis-conception..!!
given is only that force F=k√x and it is shm..
simply SHM means..F=mx (where m is a constant but k may not be a constant above)
equating them u will get k=m√x which implies k increases with displacement x, Hence it will not be a constant or SPRING CONSTANT..!
1What is meant by 'assume that he had worked only with positive x then '